Posted by Mike on .
A certain 9V battery measures 9.0 V with no load connected to it, but only 8.5 V when connected to a radio which draws 150 mA of current.
What is the internal resistance of the battery?

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bobpursley,
.150*internalresistance=98.5

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Mike,
I see you're basically finding the voltage drop that occurs as a result of that internal resistance.
Now they ask, in the second part .... What would be the terminal voltage for a 50W load? 
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Mike,
Oops. question should say 50 ohm load NOT 50W load. Sorry.
Terminal Voltage as I understand it would be the Vt = Vo  I(Ri)
Where Vt is terminal voltage, Vo is battery voltage, I is current and Ri is internal resistance. Is that correct? How would they do it? 
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bobpursley,
current = 9.0/(internal resisitance + 50)
then
terminal voltage= 9.0current*Ri 
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Mike,
thank you very much

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Mike,
sorry here's another question ... they also ask what is the maximum amount of power you can get out of this battery?
You have P = IV. Max voltage is 9V, and max current is 2.7A. Then in that case that would be 9 * 2.7 = 24.3. Did I do that right? 
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bobpursley,
No, not at all. Max power is delived when load resistance is equal to internal resistance.
Proof: Load power= Rl*Il^2=Rl(9/(Ri+Rl)^2
take the derivative:
9/(Ri+Rl)^2  18Rl/(Ri+Rl)^3
set that equal to zero, and solve for Rl
Rl=1/2 (Ri+Rl)
Rl=Ri for max power. 
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Damon,
Yes, you short circuited the terminals. In all probability the battery will not last long but it sure will get hot.

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Damon,
However as far as useful power out of the battery instead of generating heat inside you had best look for where the external voltage times the current ins maximum.